最佳拟合算法,可在回合中均匀放置对决
本文关键字:对决 算法 拟合 最佳 | 更新日期: 2023-09-27 18:35:05
我需要一种算法,它可以接受任意数量的匹配,如下所示,并按回合将它们平均分组,如果可能的话,每个人都参与一次。我按回合生成了以下 8 支和 3 支球队的比赛。我在填满我的回合时遇到了问题,并且留下了一个无法进入最后一轮的孤儿游戏。
现在回合是任意的,但正如你所知道的,每个参与者都可以在每一轮中找到(1,2,3,4,5,6,7,8)。现在这些对决可以删除或添加,并在下面生成后随机排序,因此需要均匀分布它们并在以后找到回合,我只是无法保存原始回合,因为游戏可以添加/更改/删除。
此算法应该是通用的,并且每次都进行最佳拟合。如果每支球队的对决数量不均匀,那么与其他球队相比,他们可能有额外的回合,这也需要考虑在内。这也需要高性能。
我正在寻找一些 C# .NET 或其他语言中的伪代码来了解如何完成此操作。
8支球队,每支球队10场比赛
Round 1
1 vs 2
3 vs 4
5 vs 6
7 vs 8
Round 2
1 vs 3
2 vs 4
5 vs 7
6 vs 8
Round 3
1 vs 4
2 vs 3
5 vs 8
6 vs 7
Round 4
1 vs 5
2 vs 6
3 vs 7
4 vs 8
Round 5
1 vs 6
2 vs 5
3 vs 8
4 vs 7
Round 6
1 vs 7
2 vs 8
3 vs 5
4 vs 6
Round 7
1 vs 8
2 vs 7
3 vs 6
4 vs 5
Round 8
1 vs 2
3 vs 4
5 vs 6
7 vs 8
Round 9
1 vs 3
2 vs 4
5 vs 7
6 vs 8
Round 10
1 vs 4
2 vs 3
5 vs 8
6 vs 7
3支球队,每支球队2场比赛
Round 1
1 vs 2
Round 2
2 vs 3
Round 3
1 vs 3
如果需要更具体,则必须自定义代码。
班级团队:
class Team
{
string name = "";
List<Team> playedAgainst = new List<Team>();
public string Name
{
get { return name; }
set { name = value; }
}
public Team(string name)
{
this.name = name;
}
public void AddOpponent(Team opponent)
{
this.playedAgainst.Add(opponent);
}
public bool hasPlayedAgainst(Team opponent)
{
return playedAgainst.Contains(opponent);
}
public void Reset()
{
playedAgainst.Clear();
}
public override bool Equals(object obj)
{
if(!(obj is Team))
return base.Equals(obj);
Team t = (Team)obj;
return t.name == name;
}
public override string ToString()
{
return name;
}
}
班级团队对决:
class TeamMatchup
{
List<Team> involvedTeams = new List<Team>();
List<List<Team[]>> rounds = new List<List<Team[]>>();
public void AddTeam(Team team)
{
involvedTeams.Add(team);
}
public void GenerateBattleRounds()
{
rounds = new List<List<Team[]>>();
while(true)
{
List<Team[]> round = new List<Team[]>();
foreach (Team team in involvedTeams)
{
if (!round.TrueForAll(battle => !battle.Contains(team)))
continue;
Team team2 = involvedTeams.FirstOrDefault(t => t != team && !t.hasPlayedAgainst(team) && round.TrueForAll(battle => !battle.Contains(t)));
if (team2 == null) //even count of teams
continue;
team.AddOpponent(team2);
team2.AddOpponent(team);
round.Add(new Team[] { team, team2 });
}
if (round.Count == 0)
break;
rounds.Add(round);
}
}
public override string ToString()
{
StringBuilder sb = new StringBuilder();
for (int i = 0; i < rounds.Count; i++)
{
sb.AppendLine("Round " + (i + 1));
foreach (Team[] battle in rounds[i])
{
sb.AppendLine(battle[0] + " - " + battle[1]);
}
}
return sb.ToString();
}
}
用法:
TeamMatchup matchup = new TeamMatchup();
matchup.AddTeam(new Team("Team 1"));
matchup.AddTeam(new Team("Team 2"));
matchup.AddTeam(new Team("Team 3"));
matchup.AddTeam(new Team("Team 4"));
matchup.AddTeam(new Team("Team 5"));
matchup.AddTeam(new Team("Team 6"));
matchup.AddTeam(new Team("Team 7"));
matchup.AddTeam(new Team("Team 8"));
matchup.GenerateBattleRounds();
textBox1.Text = matchup.ToString();
输出:
Round 1
Team 1 - Team 2
Team 3 - Team 4
Team 5 - Team 6
Team 7 - Team 8
Round 2
Team 1 - Team 3
Team 2 - Team 4
Team 5 - Team 7
Team 6 - Team 8
Round 3
Team 1 - Team 4
Team 2 - Team 3
Team 5 - Team 8
Team 6 - Team 7
Round 4
Team 1 - Team 5
Team 2 - Team 6
Team 3 - Team 7
Team 4 - Team 8
Round 5
Team 1 - Team 6
Team 2 - Team 5
Team 3 - Team 8
Team 4 - Team 7
Round 6
Team 1 - Team 7
Team 2 - Team 8
Team 3 - Team 5
Team 4 - Team 6
Round 7
Team 1 - Team 8
Team 2 - Team 7
Team 3 - Team 6
Team 4 - Team 5
编辑:
谁以及为什么你投票否决这个解决方案?这是对用户目标的第一个提示。如果提问者想要更详细的答案,他必须问更具体(或至少使用一个问号)。